Respuesta :
Answer:
a) σ_95% = [ ± 0.09934 ]
b) Increase the number of samples taken from each clinic and find their average over course of months for the entire year.
Step-by-step explanation:
Solution:-
- The data from the health provider for the number of CAT scans performed in a month in its clinics.
- The reported variable "X" : number of CAT scans each month expressed as the number of CAT scans per thousand members of the health plan.
- The statistical results were obtained as follows:
2.31, 2.09, 2.36, 1.95, 1.98, 2.25, 2.16, 2.07, 1.88, 1.94, 1.97,2.02
- We see that a samples were taken from n = 12 clinics were taken. The sample mean ( x_bar ) can be calculated from the following descriptive stats formula:
[tex]x_b_a_r = \frac{Sum ( x_i )}{n}\\\\x_b_a_r=\frac{2.31+2.09+2.36+1.95+1.98+2.25+2.16+2.07+1.88+1.94+1.97+2.02}{12}\\\\x_b_a_r=\frac{24.98}{12}\\\\x_b_a_r=2.0817\\[/tex]
- Similarly, we will compute the sample standard deviation ( s ) ( Normally distributed population ) for unknown population standard deviation ( σ ), the following formula is used:
[tex]s = \sqrt{\frac{Sum (x_i - x_b_a_r)^2}{n-1} } \\\\Sum (x_i - x_b_a_r)^2 = (2.31 - 2.0817)^2 + (2.09 - 2.0817)^2 + (2.36 - 2.0817)^2 + (1.95 -\\\\2.0817)^2 + (1.98 - 2.0817)^2+ (2.25 - 2.0817)^2+ (2.16 - 2.0817)^2 + (2.07 - 2.0817)^2\\\\(1.88 - 2.0817)^2 + (1.94 - 2.0817)^2 + (1.97 - 2.0817)^2 + (2.02 - 2.0817)^2\\\\= 0.26896668\\\\\\s = \sqrt{\frac{0.26896668}{12-1} } = 0.15636[/tex]
- We have two parameters ( x_bar and s ) for the approximated normal distribution of random variable X with unknown population:
x_bar = 2.0817
s = 0.15636
- The sample size n = 12 ≤ 30 and unknown population standard deviation standard normal distribution is not applicable. In such case we use t-distributions for test statistics and rejection criteria.
degree of freedom = n - 1 = 12 - 1 = 11
CI (two sided ) = 95%
Significance Level (α) = 1 - CI = 1-0.95 = 0.05
- To determine the t-critical value defined by the significance level of two sided test we have:
t-critical = t_α/2 = t _ 0.025
t-critical = t _ 0.025 = ±2.201
- We will now construct a 95% confidence interval for the population standard deviation ( σ ).
σ_95% = [ -t-critical * ( s / √n ) , t-critical * ( s / √n ) ]
σ_95% = [ -2.201 * ( 0.15636 / √12 ) , 2.201 * ( 0.15636 / √12 ) ]
σ_95% = [ ± 0.09934 ]
- To address any reservation to the standard error in standard deviation can be curtailed by increasing the sample size ( n ) at-least enough for the distribution "X" to assume standard normality. n ≥ 30
